At line 5 we enumerate all items of a collection created with list comprehension, i.e. the numbers between 1 and 1000.

Inside this loop we have another loop, but in this case we are interested in all numbers bigger than the current item of the outer collection, i.e. all numbers between a and 1000.

Then, we filter all couples (a, b) keeping only those for which we have: a² + b² = (1000 - a - b)².

According to the text of the exercise, there exists exactly one Pythagorean triplet that satisfies the equation a + b + c = 1000, so we can apply the Seq.hd (head) function to take the first (and only) element of the generated sequence.

At lines 9, 10 and 11 we evaluate a, b and c and finally we multiply them to get the answer.